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A man is a known heterozygous carrier of disease Y, which is an autosomal recessive condition. Disease Y has a prevalence of 4% in the population.
What are the man's chances of mating with a heterozygous carrier female and having a child affected with disease Y?
A. 2/25
B. 3/50
C. 1/100
D. 3/100
E. 1/400
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accor to hardy
Q2=4/100
Q=2/10
SO=P=1-2/10=8/10
CARRIER frequency=2pq=2*8/10*2/10=32/100
if he s marri with her,chance s --they will hav child with Y"diseas=32/100*1/2*1/2*1/4=2/100....................
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it will be 32/100 x 1/2 x 1/2=8/100=2/25
prob that she is a cariier=1
prob that he is a carrier =32/100 ( drdkc is right)
prob that she gives the recessive gene to a child =1/2
prob that he gives the recessive gene to a child= 1/2
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u r rt..
thanks bededoktorem..
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yes bededoktorem, you are right
